Given that \( \tan \theta=\frac{6}{5}, \pi<\theta<\frac{3 \pi}{2} \), find the exact value of each of the following. \( \begin{array}{lllll}\text { (a) } \sin (2 \theta) & \text { (b) } \cos (2 \theta) & \text { (c) } \sin \frac{\theta}{2} & \text { (d) } \cos \frac{\theta}{2} & \text { (e) } \tan (2 \theta) \\ \text { (f) } \tan \frac{\theta}{2}\end{array} \)

To solve the problem, we start with the known condition \( \tan \theta = \frac{6}{5} \) and the range \( \pi < \theta < \frac{3\pi}{2} \). This indicates that \( \theta \) is in the third quadrant, where sine is negative and cosine is also negative. ### Step 1: Find \( \sin \theta \) and \( \cos \theta \) Using the identity for tangent: \[ \tan \theta = \frac{\sin \theta}{\cos \theta} \] we can represent \( \sin \theta \) and \( \cos \theta \) in terms of a right triangle. Let the opposite side be 6 and the adjacent side be 5. The hypotenuse \( r \) can be calculated using the Pythagorean theorem: \[ r = \sqrt{6^2 + 5^2} = \sqrt{36 + 25} = \sqrt{61} \] Now we can find \( \sin \theta \) and \( \cos \theta \): \[ \sin \theta = -\frac{6}{\sqrt{61}}, \quad \cos \theta = -\frac{5}{\sqrt{61}} \] ### Step 2: Calculate \( \sin(2\theta) \) and \( \cos(2\theta) \) Using the double angle formulas: \[ \sin(2\theta) = 2 \sin \theta \cos \theta \] \[ \cos(2\theta) = \cos^2 \theta - \sin^2 \theta \] Calculating \( \sin(2\theta) \): \[ \sin(2\theta) = 2 \left(-\frac{6}{\sqrt{61}}\right) \left(-\frac{5}{\sqrt{61}}\right) = 2 \cdot \frac{30}{61} = \frac{60}{61} \] Calculating \( \cos(2\theta) \): \[ \cos(2\theta) = \left(-\frac{5}{\sqrt{61}}\right)^2 - \left(-\frac{6}{\sqrt{61}}\right)^2 = \frac{25}{61} - \frac{36}{61} = -\frac{11}{61} \] ### Step 3: Calculate \( \tan(2\theta) \) Using the formula: \[ \tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)} = \frac{\frac{60}{61}}{-\frac{11}{61}} = -\frac{60}{11} \] ### Step 4: Calculate \( \sin\left(\frac{\theta}{2}\right) \) and \( \cos\left(\frac{\theta}{2}\right) \) Using the half-angle formulas: \[ \sin\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 - \cos \theta}{2}}, \quad \cos\left(\frac{\theta}{2}\right) = \pm \sqrt{\frac{1 + \cos \theta}{2}} \] Since \( \theta \) is in the third quadrant, \( \frac{\theta}{2} \) will be in the second quadrant, where sine is positive and cosine is negative. Calculating \( \sin\left(\frac{\theta}{2}\right) \): \[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \left(-\frac{5}{\sqrt{61}}\right)}{2}} = \sqrt{\frac{1 + \frac{5}{\sqrt{61}}}{2}} = \sqrt{\frac{\frac{\sqrt{61} + 5}{\sqrt{61}}}{2}} = \sqrt{\frac{\sqrt{61} + 5}{2\sqrt{61}}} \] Calculating \( \cos\left(\frac{\theta}{2}\right) \): \[ \cos\left(\frac{\theta}{2}\right) = -\sqrt{\frac{1 + \left(-\frac{5}{\sqrt{61}}\right)}{2}} = -\sqrt{\frac{1 - \frac{5}{\sqrt{61}}}{2}} = -\sqrt{\frac{\frac{\sqrt{61} - 5}{\sqrt{61}}}{2}} = -\sqrt{\frac{\sqrt{61} - 5}{2\sqrt{61}}} \] ### Step 5: Calculate \( \tan\left(\frac{\theta}{2}\right) \) Using the formula: \[ \tan\left(\frac{\theta}{2}\right) = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \] Substituting the values: \[ \tan\left(\frac{\theta}{2}\right) = -\frac{\sqrt{\frac{\sqrt{61} + 5}{2\sqrt{61}}}}{\sqrt{\frac{\sqrt{61} - 5}{2\sqrt{61}}}} = -\sqrt{\frac{\sqrt{61} + 5}{\sqrt{61} - 5}} \] ### Final Answers Now we can summarize the results: - (a) \( \sin(2\theta) = \frac{60}{61} \) - (b) \( \cos(2\theta) = -\frac{11}{61} \) - (c) \( \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\sqrt{61} + 5}{2\sqrt{61}}} \) - (d) \( \cos\left(\frac{\theta}{2}\right) = -\sqrt{\frac{\sqrt{61} - 5}{2\sqrt{61}}} \) - (e) \( \tan(2\theta) = -\frac{60}{11} \) - (f) \( \tan\left(\frac{\theta}{2}\right) = -\sqrt{\frac{\sqrt{61} + 5}{\sqrt{61} - 5}} \)